Question

Some parts of California are particularly earthquake-prone. Suppose that in one metropolitan area, 31% of all homeowners are insured against earthquake damage. Four homeowners are to be selected at random. Let X denote the number among the four who have earthquake insurance.

(a) Find the probability distribution of X. [Hint: Let S denote a homeowner that has insurance and F one who does not. Then one possible outcome is SFSS, with probability (0.31)(0.69)(0.31)(0.31) and associated X value 3. There are 15 other outcomes.] (Round your answers to four decimal places.)
(b) What is the most likely value for X?
(c) What is the probability that at least two of the four selected have earthquake insurance? (Round your answer to four decimal places.)

Answer 1

a.) 0.0822

b.) 1

c.) 0.3659

Explanation:

Probability distribution formula is often denoted by :

P(X=r) = nCr × p^r × q^n-r

Where n = total number of samples

r = number of successful outcome of sample

p = probability of success

q = probability of failure.

If we take 4 samples,

3 of this 4 samples are successful

the success rate =S= 31% = 0.31

Failure rate = F= 0.69

a.) Then probability distribution of X becomes:

P(X=3) = 4C3 × 0.31³ × 0.69¹

P(X=3) = 0.0822 (4d.p),

b. Most likely value of X = expected value = np

= 4 × 0.31

= 1.24 ≈ 1

c.) probability that at least 2 out of the 4 have insurance = Probability that 2 have insurance) + probability that 3 have insurance + probability that 4 have insurance.

P(X=2) = 4C2 × 0.31² × 0.69² = 0.2745

P(X=3), as calculated earlier = 0.0822

P(X=4) = 4C4 × 0.31^4 × 0.69^0 = 0.0092

Total probability of having at least 2 out of those 4 insured = 0.2745 + 0.0822 + 0.0092 =0.3659.